Conditional expectation of i.i.d random variables consider the sum of random variables $Q_k=\sum_k R_k $,  $R_k$ are i.i.d.
Now I want to calculate:
$$E(Q_j| Y_{k+j}=n) =j \frac{k}{k+j}$$
 A: Since $X_i$'s are i.i.d. it follows that $c=E(X_i|Y_{n+m}=n)$ is independent of $i$. Summing over $i \leq n+m$ we get $(m+n) c=n$ so $c =\frac n {n+m}$. Finally the given expectation is $mc=\frac {mn} {m+n}$. 
A: Note that all permutations of the $X_j$ are equally likely. Thus the desired expectation is invariant if you average over them. If the sum of the first $m+n$ of the $X_j$ is $n$ (or in fact any other value, there’s no reason why it should be linked to the indices), then since each of these $m+n$ occurs in the first $m$ places in a fraction $\frac m{m+n}$ of the permutations, the expectation of the first $m$ is $\frac m{m+n}$ times that value.
A: Just another solution:
$$E(Y_m| Y_{m+n}=n)=E(\sum_{k=1}^{m} X_k| \sum_{i=1}^{m+n} X_i=n)$$
$$=\sum_{k=1}^{m}E( X_k| \sum_{i=1}^{m+n} X_i=n)=\sum_{i=1}^{m} \frac{n}{n+m}= m \frac{n}{n+m} $$
It is enough to show 
$E( X_k| \sum_{i=1}^{m+n} X_i=n)=\frac{n}{n+m}$
$$E(  \sum_{k=1}^{m+n} X_k| \sum_{i=1}^{m+n} X_i=n)=n$$
$$\Leftrightarrow$$
$$\sum_{k=1}^{m+n}E(   X_k| \sum_{i=1}^{m+n} X_i=n)=n$$ 
$$\Leftrightarrow$$
$$(m+n)E(   X_k| \sum_{i=1}^{m+n} X_i=n)=n$$ 
A: Just another  solution
lets $E(X_i)=\mu$. In non-parametric family, 
$\bar{X}$ is  sufficient and complete estimator for $\mu$(that coincide with this example).  
$$E(X_k|\sum_{i=1}^{m+n} X_i)=E(X_k|\frac{\sum_{i=1}^{m+n} X_i}{m+n})=E(X_k|\bar{X}_{(m+n)})=g(\bar{X}_{(m+n)})$$
I want to show $g(\bar{X}_{(m+n)})=\bar{X}_{(m+n)}$ almost surely.
$$E(g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)})=E(E(X_k|\bar{X}_{(m+n)}))-\mu=\mu-\mu=0$$  since $\bar{X}_{(m+n)}$ is complete and sufficient and
$g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)}$ is a function of $\bar{X}_{(m+n)}$
 so $$P(g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)}=0)=1$$
so $g(\bar{X}_{(m+n)})=\bar{X}_{(m+n)}$ almost surely. so
$$E(X_k|\sum_{i=1}^{m+n} X_i)=\bar{X}_{(m+n)}$$ and
$$E(X_k|\sum_{i=1}^{m+n} X_i=n)=E(X_k|\frac{\sum_{i=1}^{m+n} X_i}{m+n}=\frac{n}{m+n})=\frac{n}{m+n}$$. Finally
$$E(\sum_{k=1}^{m} X_k| \sum_{i=1}^{m+n} X_i=n)=m\frac{n}{m+n}$$
